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Question Number 214750 by universe last updated on 18/Dec/24

Answered by aleks041103 last updated on 21/Dec/24

J=∫_0 ^∞ ∫_z ^∞ ∫_z ^∞ cos(x^2 −y^2 )dxdydz cos(x^2 −y^2 )=cos(x^2 )cos(y^2 )+sin(x^2 )sin(y^2 ) ⇒∫_z ^∞ ∫_z ^∞ cos(x^2 −y^2 )dxdy= =∫_z ^∞ ∫_z ^∞ [cos(x^2 )cos(y^2 )+sin(x^2 )sin(y^2 )]dxdy= =A^2 +B^2 where A=∫_z ^∞ cos(x^2 )dx=(√(π/2))∫_z ^( ∞) cos((π/2)((√(2/π))x)^2 )d((√(2/π))x) B=∫_z ^∞ sin(x^2 )dx=(√(π/2))∫_z ^( ∞) sin((π/2)((√(2/π))x)^2 )d((√(2/π))x) we know C(t):=∫_0 ^( t) cos(((πx^2 )/2))dx S(t):=∫_0 ^( t) cos(((πx^2 )/2))dx where S and C are Fresnel integrals ⇒A=(√(π/2))[C(∞)−C((√(2/π))z)] ⇒B=(√(π/2))[S(∞)−S((√(2/π))z)] it is well known that C(∞)=S(∞)=(1/2) ⇒A=(√(π/2))[(1/2)−C((√(2/π))z)] ⇒B=(√(π/2))[(1/2)−S((√(2/π))z)] ⇒A^2 +B^2 =(π/2)[(1/2)+S^2 ((√(2/π))z)+C^2 ((√(2/π))z)−S((√(2/π))z)−C((√(2/π))z)]=f((√(2/π))z) ⇒J=∫_0 ^∞ f((√(2/π))z)dz= =(√(π/2))∫_0 ^∞ f(z)dz ⇒J=((π/2))^(3/2) ∫_0 ^∞ [((1/2)−C(z))^2 +((1/2)−S(z))^2 ]dz later we will show: ∫_0 ^∞ ((1/2)−C(z))^2 dz=(1/(2(√2)π)) ∫_0 ^∞ ((1/2)−S(z))^2 dz=((4−(√2))/(4π)) ⇒J=(1−((3(√2))/4))(√(π/8))

$${J}=\int_{\mathrm{0}} ^{\infty} \int_{{z}} ^{\infty} \int_{{z}} ^{\infty} {cos}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdydz} \\ $$$${cos}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)={cos}\left({x}^{\mathrm{2}} \right){cos}\left({y}^{\mathrm{2}} \right)+{sin}\left({x}^{\mathrm{2}} \right){sin}\left({y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\int_{{z}} ^{\infty} \int_{{z}} ^{\infty} {cos}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){dxdy}= \\ $$$$=\int_{{z}} ^{\infty} \int_{{z}} ^{\infty} \left[{cos}\left({x}^{\mathrm{2}} \right){cos}\left({y}^{\mathrm{2}} \right)+{sin}\left({x}^{\mathrm{2}} \right){sin}\left({y}^{\mathrm{2}} \right)\right]{dxdy}= \\ $$$$={A}^{\mathrm{2}} +{B}^{\mathrm{2}} \\ $$$${where}\: \\ $$$${A}=\int_{{z}} ^{\infty} {cos}\left({x}^{\mathrm{2}} \right){dx}=\sqrt{\frac{\pi}{\mathrm{2}}}\int_{{z}} ^{\:\infty} {cos}\left(\frac{\pi}{\mathrm{2}}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{x}\right)^{\mathrm{2}} \right){d}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{x}\right) \\ $$$${B}=\int_{{z}} ^{\infty} {sin}\left({x}^{\mathrm{2}} \right){dx}=\sqrt{\frac{\pi}{\mathrm{2}}}\int_{{z}} ^{\:\infty} {sin}\left(\frac{\pi}{\mathrm{2}}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{x}\right)^{\mathrm{2}} \right){d}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{x}\right) \\ $$$${we}\:{know} \\ $$$${C}\left({t}\right):=\int_{\mathrm{0}} ^{\:{t}} {cos}\left(\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right){dx} \\ $$$${S}\left({t}\right):=\int_{\mathrm{0}} ^{\:{t}} {cos}\left(\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right){dx} \\ $$$${where}\:{S}\:{and}\:{C}\:{are}\:{Fresnel}\:{integrals} \\ $$$$\Rightarrow{A}=\sqrt{\frac{\pi}{\mathrm{2}}}\left[{C}\left(\infty\right)−{C}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{z}\right)\right] \\ $$$$\Rightarrow{B}=\sqrt{\frac{\pi}{\mathrm{2}}}\left[{S}\left(\infty\right)−{S}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{z}\right)\right] \\ $$$${it}\:{is}\:{well}\:{known}\:{that} \\ $$$${C}\left(\infty\right)={S}\left(\infty\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{A}=\sqrt{\frac{\pi}{\mathrm{2}}}\left[\frac{\mathrm{1}}{\mathrm{2}}−{C}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{z}\right)\right] \\ $$$$\Rightarrow{B}=\sqrt{\frac{\pi}{\mathrm{2}}}\left[\frac{\mathrm{1}}{\mathrm{2}}−{S}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{z}\right)\right] \\ $$$$ \\ $$$$\Rightarrow{A}^{\mathrm{2}} +{B}^{\mathrm{2}} =\frac{\pi}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}+{S}^{\mathrm{2}} \left(\sqrt{\frac{\mathrm{2}}{\pi}}{z}\right)+{C}^{\mathrm{2}} \left(\sqrt{\frac{\mathrm{2}}{\pi}}{z}\right)−{S}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{z}\right)−{C}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{z}\right)\right]={f}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{z}\right) \\ $$$$\Rightarrow{J}=\int_{\mathrm{0}} ^{\infty} {f}\left(\sqrt{\frac{\mathrm{2}}{\pi}}{z}\right){dz}= \\ $$$$=\sqrt{\frac{\pi}{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} {f}\left({z}\right){dz} \\ $$$$\Rightarrow{J}=\left(\frac{\pi}{\mathrm{2}}\right)^{\mathrm{3}/\mathrm{2}} \int_{\mathrm{0}} ^{\infty} \left[\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right)^{\mathrm{2}} \right]{dz} \\ $$$${later}\:{we}\:{will}\:{show}: \\ $$$$\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right)^{\mathrm{2}} {dz}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\pi} \\ $$$$\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right)^{\mathrm{2}} {dz}=\frac{\mathrm{4}−\sqrt{\mathrm{2}}}{\mathrm{4}\pi} \\ $$$$\Rightarrow{J}=\left(\mathrm{1}−\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}\right)\sqrt{\frac{\pi}{\mathrm{8}}} \\ $$

Commented by aleks041103 last updated on 21/Dec/24

∫((1/2)−C(z))^2 dz= =z((1/2)−C(z))^2 +2∫z((1/2)−C(z))C ′(z)dz ∫z((1/2)−C(z))C ′(z)= =∫zcos((π/2)z^2 )((1/2)−C(z))dz= =(1/π)∫((1/2)−C(z))d(sin(((πz^2 )/2)))= =(1/π)sin(((πz^2 )/2))((1/2)−C(z))+(1/π)∫sin(((πz^2 )/2))C ′(z)dz ∫sin(((πz^2 )/2))C ′(z)dz=∫sin(((πz^2 )/2))cos(((πz^2 )/2))dz= =(1/(2(√2)))∫sin(((π((√2)z)^2 )/2))d((√2)z)=(1/(2(√2)))S((√2)z) ⇒∫z((1/2)−C(z))C ′(z)= =(1/π)sin(((πz^2 )/2))((1/2)−C(z))+((S((√2)z))/(2(√2)π)) ⇒F(z)=∫((1/2)−C(z))^2 dz= =z((1/2)−C(z))^2 +(2/π)sin(((πz^2 )/2))((1/2)−C(z))+((S((√2)z))/( (√2)π)) F(0)=0 F(∞)=((S(∞))/( (√2)π))=(1/(2(√2)π)) ⇒∫_0 ^∞ ((1/2)−C(z))^2 dz=(1/(2(√2)π))

$$\int\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right)^{\mathrm{2}} {dz}= \\ $$$$={z}\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right)^{\mathrm{2}} +\mathrm{2}\int{z}\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right){C}\:'\left({z}\right){dz} \\ $$$$\int{z}\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right){C}\:'\left({z}\right)= \\ $$$$=\int{zcos}\left(\frac{\pi}{\mathrm{2}}{z}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right){dz}= \\ $$$$=\frac{\mathrm{1}}{\pi}\int\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right){d}\left({sin}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right)\right)= \\ $$$$=\frac{\mathrm{1}}{\pi}{sin}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right)+\frac{\mathrm{1}}{\pi}\int{sin}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right){C}\:'\left({z}\right){dz} \\ $$$$\int{sin}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right){C}\:'\left({z}\right){dz}=\int{sin}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right){cos}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right){dz}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int{sin}\left(\frac{\pi\left(\sqrt{\mathrm{2}}{z}\right)^{\mathrm{2}} }{\mathrm{2}}\right){d}\left(\sqrt{\mathrm{2}}{z}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{S}\left(\sqrt{\mathrm{2}}{z}\right) \\ $$$$\Rightarrow\int{z}\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right){C}\:'\left({z}\right)= \\ $$$$=\frac{\mathrm{1}}{\pi}{sin}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right)+\frac{{S}\left(\sqrt{\mathrm{2}}{z}\right)}{\mathrm{2}\sqrt{\mathrm{2}}\pi} \\ $$$$\Rightarrow{F}\left({z}\right)=\int\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right)^{\mathrm{2}} {dz}= \\ $$$$={z}\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right)^{\mathrm{2}} +\frac{\mathrm{2}}{\pi}{sin}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right)+\frac{{S}\left(\sqrt{\mathrm{2}}{z}\right)}{\:\sqrt{\mathrm{2}}\pi} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${F}\left(\infty\right)=\frac{{S}\left(\infty\right)}{\:\sqrt{\mathrm{2}}\pi}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\pi} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}−{C}\left({z}\right)\right)^{\mathrm{2}} {dz}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\pi} \\ $$

Commented by aleks041103 last updated on 21/Dec/24

∫((1/2)−S(z))^2 dz= =z((1/2)−S(z))^2 +2∫z((1/2)−S(z))S ′(z)dz ∫z((1/2)−S(z))S ′(z)dz= =∫z sin(((πz^2 )/2))((1/2)−S(z))dz= =−(1/π)∫((1/2)−S(z))d(cos(((πz^2 )/2)))= =−(1/π)cos(((πz^2 )/2))((1/2)−S(z))−(1/π)∫cos(((πz^2 )/2))S ′(z)dz ∫cos(((πz^2 )/2))S ′(z)dz= =∫cos(((πz^2 )/2))sin(((πz^2 )/2))dz= =(1/(2(√2)))∫sin(((π((√2)z)^2 )/2))d((√2)z)= =((S((√2)z))/(2(√2))) ⇒∫z((1/2)−S(z))S ′(z)dz= =−(1/π)cos(((πz^2 )/2))((1/2)−S(z))−((S((√2)z))/(2(√2)π)) ⇒G(z)=∫((1/2)−S(z))^2 dz= =z((1/2)−S(z))^2 −(2/π)cos(((πz^2 )/2))((1/2)−S(z))−((S((√2)z))/( (√2)π)) G(0)=−(1/π) G(∞)=−(1/(2(√2)π)) ⇒∫_0 ^( ∞) ((1/2)−S(z))^2 dz=(1/π)(1−(1/(2(√2))))=((4−(√2))/(4π))

$$\int\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right)^{\mathrm{2}} {dz}= \\ $$$$={z}\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right)^{\mathrm{2}} +\mathrm{2}\int{z}\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right){S}\:'\left({z}\right){dz} \\ $$$$\int{z}\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right){S}\:'\left({z}\right){dz}= \\ $$$$=\int{z}\:{sin}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right){dz}= \\ $$$$=−\frac{\mathrm{1}}{\pi}\int\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right){d}\left({cos}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right)\right)= \\ $$$$=−\frac{\mathrm{1}}{\pi}{cos}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right)−\frac{\mathrm{1}}{\pi}\int{cos}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right){S}\:'\left({z}\right){dz} \\ $$$$\int{cos}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right){S}\:'\left({z}\right){dz}= \\ $$$$=\int{cos}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right){sin}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right){dz}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int{sin}\left(\frac{\pi\left(\sqrt{\mathrm{2}}{z}\right)^{\mathrm{2}} }{\mathrm{2}}\right){d}\left(\sqrt{\mathrm{2}}{z}\right)= \\ $$$$=\frac{{S}\left(\sqrt{\mathrm{2}}{z}\right)}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow\int{z}\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right){S}\:'\left({z}\right){dz}= \\ $$$$=−\frac{\mathrm{1}}{\pi}{cos}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right)−\frac{{S}\left(\sqrt{\mathrm{2}}{z}\right)}{\mathrm{2}\sqrt{\mathrm{2}}\pi} \\ $$$$\Rightarrow{G}\left({z}\right)=\int\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right)^{\mathrm{2}} {dz}= \\ $$$$={z}\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right)^{\mathrm{2}} −\frac{\mathrm{2}}{\pi}{cos}\left(\frac{\pi{z}^{\mathrm{2}} }{\mathrm{2}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right)−\frac{{S}\left(\sqrt{\mathrm{2}}{z}\right)}{\:\sqrt{\mathrm{2}}\pi} \\ $$$${G}\left(\mathrm{0}\right)=−\frac{\mathrm{1}}{\pi} \\ $$$${G}\left(\infty\right)=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\pi} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\infty} \left(\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({z}\right)\right)^{\mathrm{2}} {dz}=\frac{\mathrm{1}}{\pi}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)=\frac{\mathrm{4}−\sqrt{\mathrm{2}}}{\mathrm{4}\pi} \\ $$

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